[AntonioLao]

Three necessary and sufficient conditions for rigidity of a particle are the following: (r2-r1)(v2-v1) = 0, (r3-r2)(v3-v2) = 0, and (r3-r1)(v3-v1) = 0. Applying the mass-time operator (p/v)(d/dt) to either the left or the right factors give the following: (p2-p1)(v2-v1) = 0, (p3-p2)(v3-v2) = 0, and (p3-p1)(v3-v1) = 0, and (r2-r1)(f2-f1) = 0, (r3-r2)(f3-f2) = 0, and (r3-r1)(f3-f1) = 0. The factors containing forces and space-time displacements could be used to describe the rigidity of the quantum vacuum. The following properties are necessary for reducing some of the factors: f1 ^ f2, f1 ^ r1, f1 ^ r2, f1 || f2, f1 || r1, f1 || r2, f2 ^ r1, f2 ^ r2, f2 || r1, f1 || r2, r1 || r2, r1 ^ r2 . Note that not all of these properties can be simultaneously satisfied.

[mkirkpatrick]

Are particles actually rigid Antonio?A rigid wave?Well prehaps if it were frozen seawater in
the Bering sea!What say you of a rigid wave,or pulse.

regards michael.

[AntonioLao]

Originally Posted by mkirkpatrick Are particles actually rigid

Rigidity is defined classical mechanically as the vanishing product of differences of radius vectors and differences of velocity vectors. For waves, their radius vectors could never be zero neither their velocity vectors which is constant for EM waves as speed of light.